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Katena32 [7]
1 year ago
8

Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi

ty has a magnitude of 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m ​ 3, point, 7, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. We want to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance.
Physics
1 answer:
Doss [256]1 year ago
5 0

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

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When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
Oxana [17]

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

m=5\cdot 10^{-3}g

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

1 g : 1000 mg = 0.005 g : x

and we find

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

4 0
2 years ago
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache
FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

      W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis  

      N - W cos θ  = 0

The equation for the force of friction is

      fr = μ N

      fr = μ mg cos θ

We replace at work

     W = - μ m g cos θ  d

Mechanical energy in the lower part of the embankment

      Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

     Em_{f} = U = m g y

     W = ΔEm =  Em_{f} - Em₀

    - μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

     sin θ = y / d

     y = d sin θ

   

    - μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

     d (g syn θ + μ g cos θ) = ½ v²

     d = v²/2 g (sintea + myy cos tee)

     d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

     d = 1555.85 /7.8145

     d = 199.1 m

Let's use trigonometry to find the height

      sin 16 = y / d

      y = d sin 16

      y = 199.1 sin 16

      y = 54.9 m

8 0
2 years ago
An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic
nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0   →   <em>n</em> = <em>mg</em> = 1876 N

where <em>n</em> is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0   →   <em>f</em> = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that

<em>f</em> = <em>µn</em>   →   <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

8 0
2 years ago
Which is found farthest from the center of an atom?
xxTIMURxx [149]
Electron

Hope this helps
5 0
1 year ago
Your film idea is about drones that take over the world. In the script, two drones are flying horizontally at the same speed and
Stella [2.4K]

Answer:

vₓ = 20 m/s,    v_{y}  = -15 m / s

Explanation:

This is a conservation of moment problem, since it is a vector quantity we can work each axis independently

The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved

X axis

Initial moment. Before the crash

         p₀ = m₁ v₀ₓ + m₂ v₀ₓ

Final moment. After the crash

       p_{fx} = (m₁ + m₂) vₓ

      p₀ₓ = p_{fx}

      m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ

       vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)

       vₓ = v₀ₓ  = 20 m/s

Y Axis

Initial

         p_{oy} = m₁ v_{oy}

Final

         p_{fy} = (m₁ + m₂) v_{y}

         p_{oy} = p_{fy}

the drom rises and when it falls it has the same speed because there is no friction    v_{oy} = -60 m/s          

 

           m₁ v_{oy} = (m₁ + m₂) v_{y}

            v_{y} = m₁ / (m₁ + m₂) v_{oy}

            v_{y}  = 1/4    60

            v_{y}  = -15 m / s

Vertical speed is down

5 0
2 years ago
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